GRE_Math_Bible_eBook

A

B

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Very Hard

29.Two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 2, respectively. A new alloy X is formed by mixing the two alloys A and B in the ratio 4 : 3. What is the ratio of the composition of the two basic elements in alloy X ?

(A)1 : 1

(B)2 : 3

(C)5 : 2

(D)4 : 3

(E)7 : 9

294GRE Math Bible

30.Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce, making a profit of 20 percent. What was the ratio of the mixture?

(A)1 : 10

(B)1 : 5

(C)2 : 7

(D)3 : 8

(E)5 : 7

31.The Savings of an employee equals Income minus Expenditure. If their Incomes ratio is 1 : 2 : 3 and their Expenses ratio is 3 : 2 : 1, then what is the order of the employees A, B, and C in the increasing order of the size of their savings?

(A)A > B > C

(B)A > C > B

(C)B > A > C

(D)B > C > A

(E)C > B > A

32.A cask initially contains pure alcohol up to the brim. The cask can be emptied by removing exactly 5 liters at a time . Each time this is done, the cask must be filled back to the brim with water. The capacity of the cask is 15 liters. When the cask is completely emptied and filled back to the brim two times, what is the ratio of alcohol to water in the cask?

(A)1 : 2

(B)2 : 1

(C)1 : 1

(D)4 : 5

(E)5 : 4

Ratio & Proportion 295

Answers and Solutions to Problem Set Q

Easy

1. This problem can be solved by setting up a proportion. Note that 1 pound has 16 ounces, so 3 pounds has 48 (= 3 16) ounces. Now, the proportion, in cents to ounces, is

800 cents

48 = 9

or

cents = 9 80048 = 150

The answer is (E).

Medium

2. We know that the angle made by a line is 180°. Applying this to line m yields x + y + 90 = 180. Subtracting 90 from both sides of this equation yields x + y = 90. We are also given that x : y = 2 : 3. Hence, x/y = 2/3. Multiplying this equation by y yields x = 2y/3. Plugging this into the equation x + y = 90 yields

2y/3 + y = 90

5y/3 = 90

5y = 270 y = 54

The answer is (D).

3. Joining the midpoints of the opposite sides of the rectangle ABCD yields the following figure:

In the given figure, the bigger rectangle ABCD contains four small rectangles, each one divided by a diagonal. Since diagonals cut a rectangle into two triangles of equal area, in each of the small rectangles, the regions to either side (shaded and un-shaded) have equal area. Hence, even in the bigger rectangle, the area of the shaded and un-shaded regions are equal, so the required ratio is 1 : 1. The answer is (A).

4. We are given that r is 25% greater than s. Hence, r = (1 + 25/100)s. So, r/s = 1 + 25/100 = 1.25. The answer is (E).

296 GRE Math Bible

5. We are given that the ratio of the sum of the reciprocals of x and y to the product of the reciprocals of x and y is 1 : 3. Writing this as an equation yields

xy

x+ y

The answer is (A).

6.Let the number of chickens, pigs and horses on Richard’s farm be c, p, and h. Then forming the given ratio yields c : p : h = 33 : 17 : 21. Let c equal 33k, p equal 17k, and h equal 21k, where k is a positive integer (such that c : p : h = 33 : 17 : 21). Then the total number of pigs and horses is 17k + 21k = 38k; and the total number of pigs, horses and chickens is 17k + 21k + 33k = 71k. Hence, the required fraction equals 38k/71k = 38/71. The answer is (C).

7.Suppose Marc’s income is $300 and his boss’s income is $400 (incomes match the given ratio, 3 : 4). Also, suppose Marc’s expenditure is $100 and his boss’s expenditure is $200 (expenditures match the given ratio, 1 : 2). Then the savings from income of Marc is 300 – 100 = 200 and that of his boss is 400 – 200 =

200.In this case, Marc’s savings equals his boss’s savings.

Now, suppose the expenditures are different: Marc’s expenditure is $150 and his boss’s expenditure is $300 (expenditures match the given ratio, 1 : 2). In this case, Marc’s savings is 300 – 150 = 150, and his boss’s savings is 400 – 300 = 100 (savings are not equal). Hence, we have a double case, and the answer is (D).

8. This problem can be solved by setting up a proportion between the number of cookies and the number of cups of sugar required to make the corresponding number of cookies. Since there are 12 items in a dozen, 2-dozen cookies is 2 12 = 24 cookies. Since 3/2 cups are required to make the 24 cookies, we have the proportion

Hence, Column A equals 45/24, which is less than 2 (= Column B). Hence, Column B is greater, and the answer is (B).

9. We are given that ax : by : cz = 1 : 2 : –3. Let ax = t, by = 2t, and cz = –3t (such that ax : by : cz = 1 : 2 : –3). Then ax + by + cz = t + 2t – 3t = 0. The answer is (A).

Ratio & Proportion 297

10. Forming the ratio yields a/b = 2/3. Solving for b yields b = 3a/2. Since a is positive (given), this equation tells us that b is also positive. Hence, b + 5 is positive, and we can safely multiply both columns by b + 5. This yields

Subtracting 5 from both columns yields

Since both a and b are positive and b is one-and-a-half times larger than a (b = 3a/2), Column B is larger than Column A. The answer is (B).

Hard

11. Angles B and C in triangle ABC equal 60° and 90°, respectively. Since the sum of the angles in a triangle is 180°, the third angle of the triangle, A, must equal 180° – (60° + 90°) = 30°.

So, in the two triangles, ABC and EDF, we have A = E = 30° and C = F = 90° (showing that at least two corresponding angles are equal). So, the two triangles are similar. Hence, the ratios of the corresponding sides in the two triangles are equal. Hence, we have

12. The area of the parallelogram ABCD equals base height = DC AF = 10AF (from the figure, DC = 10).

The area of rectangle AECF equals length width = FC AF.

We are given that the ratio of the two areas is 5 : 3. Forming the ratio gives

10AF = 5

FC AF 3

10 5

FC = 3

FC = 10 35 = 6

From the figure, we have DC = DF + FC. Now, DC = 10 (from the figure) and FC = 6 (as we just calculated). So, 10 = DF + 6. Solving this equation for DF yields DF = 4. Since F is one of the angles in the rectangle AECF, it is a right angle. Since AFC is right-angled, AFD must also measure 90°. So,AFD is right-angled, and The Pythagorean Theorem yields

AF2 + DF2 = AD2

AF2 + 42 = 52

AF2 = 52 – 42 = 25 – 16 = 9

AF = 3

Now, the area of the rectangle AECF equals FC AF = 6 3 = 18. The answer is (A).

13. Let the number of girls be x. Since the ratio of the girls to boys is 3 : 4, the number of boys equals (4/3)x. Hence, the number of students in the class equals x + 4x/3 = 7x/3. We are given that 10% of girls are blue eyed, and 10% of x is 10/100 x = x/10. Also, 20% of the boys are blue eyed, and 20% of 4x/3 is (20/100)(4x/3) = 4x/15.

Hence, the total number of blue-eyed students is x/10 + 4x/15 = 11x/30.

298 GRE Math Bible

14. The cost of production is proportional to the number of units produced. Hence, we have the equation

The Cost of Production = k Quantity, where k is a constant.

We are given that 300 units cost 300 dollars. Putting this in the proportionality equation yields 300 = k 300. Solving the equation for k yields k = 300/300 = 1. Hence, the Cost of Production of 270 units equals k 270 = 1 270 = 270. The answer is (A).

15.Kelvin takes 3 minutes to inspect a car, and John takes 4 minutes to inspect a car. Hence, after t minutes, Kelvin inspects t/3 cars and John inspects t/4 cars. Hence, the ratio of the number of cars inspected by them is t/3 : t/4 = 1/3 : 1/4 = 4 : 3. The answer is (D).

16.The formula for simple interest rate (in percentage) equals PrincipalInterest 100 . From this formula, it is

clear that the principal being the same (= $1000 in the two given cases), the interest rates are directly proportional to the interests earned on them ($80 in case of Bank A and $140 in case of Bank B). Hence, the ratio of the interest rates is 80 : 140 = 4 : 7. The answer is (D).

17.We are given the equations x = 10a, y = 3b, z = 7c and the proportion x : y : z = 10 : 3 : 7.

Substituting the first three equations into the last equation (ratio equation) yields 10a : 3b : 7c = 10 : 3 : 7. Forming the resultant ratio yields 10a/10 = 3b/3 = 7c/7. Simplifying the equation yields a = b = c. Hence, both a and b equal c. Substituting the result in the given equations x = 10a, y = 3b, z = 7c yields x = 10a = 10c, y = 3b = 3c, and z = 7c. Now, we have

18. Two triangles are similar when their corresponding angle ratios or corresponding side ratios are the same.

Name the first triangle ABC and the second triangle DEF. Now, the angle ratio of the first triangle isA : B : C = 2 : 3 : 4, and angle ratio for the second triangle is D : E : F = 4 : 3 : 2, while the sum of the angles is 180 degrees.

The corresponding ordering ratio can be safely reversed and doing this for the second ratio equation yieldsF : E : D = 2 : 3 : 4, while the sum of the angles is 180 degrees.

Since the angles in triangle ABC equal the corresponding angles in triangle FED, triangles ABC and FED are similar. Hence, the answer is (A).

Ratio & Proportion 299

19. Let 1 pound of the rice of the first type ($0.8 per pound) be mixed withp pounds of the rice of the second type ($0.9 per pound). Then the total cost of the 1 + p pounds of the rice is

If this equals $0.825 per pound (given), then we have the equation

0.8+ 0.9 p = 0.825 1+ p

0.8+ 0.9p = 0.825(1 + p)

0.8+ 0.9p = 0.825 + 0.825p 0.9p – 0.825p = 0.825 – 0.8 900p – 825p = 825 – 800 75p = 25

p = 25/75 = 1/3

Hence, the proportion of the two rice types is 1 : 1/3, which also equals 3 : 1. Hence, the answer is (E).

20. We are given the equations x = a, y = 2b, z = 3c, and the proportionx : y : z = 1 : 2 : 3. Substituting the first three equations into the last equation yields a : 2b : 3c = 1 : 2 : 3. Forming the resultant ratio yields a/1 = 2b/2 = 3c/3. Simplifying the equation yields a = b = c. Thus, we have that both a and b equal c. Hence, from the given equations, we have x = a = c, y = 2b = 2c, and z = 3c.

=6c = 2 = Column B.

3c

Hence, the answer is (C).

21. Forming the given ratio yields

A/1 = B/2 = C/3 = k, for some integer A = k, B = 2k, and C = 3k

Choice (A): A + B + C = k + 2k + 3k = 6k. This is a perfect square only when k is a product of 6 and a perfect square number. For example, when k is 6 92, 6k = 62 92, a perfect square. In all other cases (suppose k = 2, then 6k = 12), it is not a perfect square. Hence, reject.

Choice (B): A2 + B2 + C2 = k2 + (2k)2 + (3k)2 = k2 + 4k2 + 9k2 = 14k2. This is surely not a perfect square. For example, suppose k equals 2. Then 14k2 = 56, which is not a perfect square. Hence, reject.

Choice (C): A3 + B3 + C3 = k3 + (2k)3 + (3k)3 = k3 + 8k3 + 27k3 = 36k3. This is a perfect square only when k3

Choice (D): 3A2 + B2 + C2 = 3k2 + (2k)2 + (3k)2 = 3k2 + 4k2 + 9k2 = 16k2 = 42k2 = (4k)2. The square root of (4k)2 is 4k and is an integer for any integer value of k. Hence, this expression must always result in a perfect square. Choose (D).

Choice (E): 3A2 + 4B2 + 4C2 = 3k2 + 4(2k)2 + 4(3k)2 = 3k2 + 16k 2 + 36k 2 = 55k2. This is surely not a perfect square. Hence, reject.

The answer is (D).

300 GRE Math Bible

22. Let x be the weight of the full stone. Then the weight of each of the three broken pieces is x/3.

Since we are given that the value of the stone is proportional to the square of its weight, we have that if kx2

x 2

is the value of the full stone, then the value of each small stone should be k , where k is the3

proportionality constant. Hence, the value of the three pieces together is

kx2

3

Hence, Column A equals kx2/3 and Column B equals kx2. Hence, Column A is one third of Column B. The answer is (B).

the answer is (B).

25. Let k and 3k be the number of cheetahs and pandas five years ago. Let d and 2d be the corresponding numbers now.

The increase in the number of cheetahs is d – k, and the increase in the number of pandas is 2d – 3k.

Now, suppose the increase in the number of cheetahs equals the increase in the number of pandas. Then we have the equation d – k = 2d – 3k. Solving the equation for d yields d = 2k. Hence, we have a case here. Suppose k = 3. Then d = 2k = 6. The case supposes there were 3 cheetahs and 9 pandas 5 years ago, and now, there are 6 cheetahs and 12 pandas. Hence, the increase is the same.

Now, suppose the increase in the number of cheetahs is less than that of the pandas. Then we have the inequality d – k < 2d – 3k. Solving the inequality yields d > 2k. Hence, suppose k = 3 and d = 7. The case refers to when there are 3 cheetahs and 9 pandas five years ago and now there are 7 cheetahs and 14 pandas.

Now, suppose the increase in the number of cheetahs is greater than that of the pandas. Then we have the inequality d – k > 2d – 3k. Solving the inequality yields d < 2k. Hence, suppose k = 3 and d = 5. The case refers to when there are 3 cheetahs and 9 pandas five years ago, and now, there are 5 cheetahs and just 10 pandas. Here, the increase in cheetahs is greater than the increase in pandas.

Hence, we have a double case, and the answer is (D).